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        <p>月饼是中国人在中秋佳节时吃的一种传统食品，不同地区有许多不同风味的月饼。现给定所有种类月饼的库存量、总售价、以及市场的最大需求量，请你计算可以获得的最大收益是多少。</p>
<p>注意：销售时允许取出一部分库存。样例给出的情形是这样的：假如我们有3种月饼，其库存量分别为18、15、10万吨，总售价分别为75、72、45亿元。如果市场的最大需求量只有20万吨，那么我们最大收益策略应该是卖出全部15万吨第2种月饼、以及5万吨第3种月饼，获得 72 + 45/2 = 94.5（亿元）。</p>
<p>输入格式：</p>
<p>每个输入包含1个测试用例。每个测试用例先给出一个不超过1000的正整数N表示月饼的种类数、以及不超过500（以万吨为单位）的正整数D表示市场最大需求量。随后一行给出N个正数表示每种月饼的库存量（以万吨为单位）；最后一行给出N个正数表示每种月饼的总售价（以亿元为单位）。数字间以空格分隔。</p>
<p>输出格式：</p>
<p>对每组测试用例，在一行中输出最大收益，以亿元为单位并精确到小数点后2位。</p>
<p>输入样例：<br>3 20<br>18 15 10<br>75 72 45<br>输出样例：<br>94.50</p>
<p>Mooncake is a Chinese bakery product traditionally eaten during the Mid-Autumn Festival. Many types of fillings and crusts can be found in traditional mooncakes according to the region’s culture. Now given the inventory amounts and the prices of all kinds of the mooncakes, together with the maximum total demand of the market, you are supposed to tell the maximum profit that can be made.</p>
<p>Note: partial inventory storage can be taken. The sample shows the following situation: given three kinds of mooncakes with inventory amounts being 180, 150, and 100 thousand tons, and the prices being 7.5, 7.2, and 4.5 billion yuans. If the market demand can be at most 200 thousand tons, the best we can do is to sell 150 thousand tons of the second kind of mooncake, and 50 thousand tons of the third kind. Hence the total profit is 7.2 + 4.5/2 = 9.45 (billion yuans).</p>
<p>Input Specification:</p>
<p>Each input file contains one test case. For each case, the first line contains 2 positive integers N (&lt;=1000), the number of different kinds of mooncakes, and D (&lt;=500 thousand tons), the maximum total demand of the market. Then the second line gives the positive inventory amounts (in thousand tons), and the third line gives the positive prices (in billion yuans) of N kinds of mooncakes. All the numbers in a line are separated by a space.</p>
<p>Output Specification:</p>
<p>For each test case, print the maximum profit (in billion yuans) in one line, accurate up to 2 decimal places.</p>
<p>Sample Input:<br>3 200<br>180 150 100<br>7.5 7.2 4.5<br>Sample Output:<br>9.45<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line">//#include &quot;math.h&quot;</span><br><span class="line">#include &quot;string.h&quot;</span><br><span class="line">#include &quot;algorithm&quot;</span><br><span class="line">using namespace std;</span><br><span class="line">struct mooncake&#123;</span><br><span class="line">    double store;//存量</span><br><span class="line">    double sell;//总销售价</span><br><span class="line">    double price;//单价</span><br><span class="line">&#125;moon[1010];</span><br><span class="line">bool cmp(mooncake a, mooncake b)&#123;</span><br><span class="line">    return a.price &gt; b.price;</span><br><span class="line">&#125;</span><br><span class="line">int main()&#123;</span><br><span class="line">    int n, d;</span><br><span class="line">    double sum = 0.0;</span><br><span class="line">    scanf(&quot;%d%d&quot;, &amp;n, &amp;d);</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        scanf(&quot;%lf&quot;, &amp;moon[i].store);</span><br><span class="line">    &#125;</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        scanf(&quot;%lf&quot;, &amp;moon[i].sell);</span><br><span class="line">        moon[i].price = moon[i].sell / moon[i].store;</span><br><span class="line">    &#125;</span><br><span class="line">    sort(moon, moon + n, cmp);</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        if (moon[i].store &lt;= d) &#123;//存量不能满足要求</span><br><span class="line">            sum += moon[i].sell;</span><br><span class="line">            d -= moon[i].store;</span><br><span class="line">        &#125;else&#123;</span><br><span class="line">            sum += moon[i].price * d ;</span><br><span class="line">            break;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    printf(&quot;%.2f\n&quot;, sum);</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>

      
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